Question 1

In semiconductor electronics, what is the primary application of a Zener diode in practical circuits, and why is it suited for that role?

Accepted Answer

Introduction / Context: Zener diodes are specially doped semiconductor diodes designed to operate in the breakdown region safely. They are ubiquitous in power supplies, reference modules, and protection networks. This question tests conceptual understanding of how Zener breakdown is used to hold a nearly constant voltage across a load despite changes in input voltage or load current. Given Data / Assumptions: The Zener diode is reverse-biased during normal operation. It exhibits a sharp breakdown at the Zener voltage Vz and maintains an approximately constant voltage there. Series resistance is used to limit current to a safe value. Concept / Approach: The key feature is the diode’s near-constant voltage in breakdown (Zener and/or avalanche mechanisms). By placing the Zener diode in shunt with the load and using a series resistor from an unregulated source, the Zener clamps the output at Vout ≈ Vz, thus regulating the supply to the load even if the source varies within limits. Step-by-Step Solution: Place Zener diode in reverse across the load; choose Vz equal to desired regulated voltage.Use a series resistor Rs from the unregulated input Vs to limit current.Design inequality: IZ(min) ≤ IZ ≤ IZ(max). Load current IL plus Zener current IZ flows through Rs.Output voltage is held at Vout ≈ Vz as long as IZ stays within its safe operating range. Verification / Alternative check: Compute Rs = (Vs(min) − Vz)/(IL(max) + IZ(min)) to ensure regulation at worst case. Check power: Pz = Vz * IZ(max). Why Other Options Are Wrong: Amplifier circuits: a Zener does not provide gain; it is a reference/clamp, not an amplifying element.Both voltage regulator and amplifier: incorrect because a Zener is not used as a linear gain device.None of the above: false since regulation is the canonical application.Precision rectifier: implemented with op-amps and diodes, not a Zener function. Common Pitfalls: Confusing forward diode drops with Zener regulation in reverse breakdown.Ignoring current and power ratings of the Zener and series resistor. Final Answer: voltage regulator circuit

Question 2

In an n–p–n bipolar junction transistor (BJT), which carriers are the majority carriers within the base region, and why?

Accepted Answer

Introduction / Context: Carrier types and doping polarities in BJTs determine device operation, current flow, and transistor gain. For an n–p–n device, understanding which carriers dominate in each region (emitter, base, collector) is fundamental to analyzing injection, recombination, and current gains (α and β). Given Data / Assumptions: An n–p–n transistor has an n-type emitter, p-type base, and n-type collector. “Majority carriers” refer to carriers introduced by the intentional doping of that region (not the injected minority carriers from another region). Concept / Approach: By definition, the base in an n–p–n BJT is p-type; therefore, holes are its majority carriers. Although the useful conduction involves electrons injected from the n-type emitter into the base (where they are minority carriers), the base remains p-type with holes as majority carriers and electrons as minority carriers within it. Step-by-Step Solution: Identify doping: base is p-type → majority carriers are holes.Recognize device action: emitter injects electrons into base; these electrons are minority carriers in the base and are partially collected by the collector.Therefore the correct answer is holes. Verification / Alternative check: Compare with a p–n–p BJT: there the base is n-type and electrons would be majority; this symmetry check confirms the reasoning. Why Other Options Are Wrong: Electrons: they are minority in the base of an n–p–n.Both or either: majority carriers are determined by doping, not by operating condition ambiguity.Ions: irrelevant to majority carrier conduction in a semiconductor device. Common Pitfalls: Confusing “majority within base” with “dominant conduction mechanism” of the overall transistor. Final Answer: holes

Question 3

For a p–n–p transistor, the emitter current IE has two components: IEp due to holes injected from p to n, and IEn due to electrons injected from n to p. Which component dominates under normal operation?

Accepted Answer

Introduction / Context: Emitter efficiency is critical to BJT performance. In a p–n–p transistor, the emitter is p-type and is heavily doped to inject holes efficiently into the base. Understanding which component of emitter current dominates clarifies why high emitter efficiency leads to high current gain and good transistor action. Given Data / Assumptions: Device type: p–n–p BJT with p-type emitter, n-type base, p-type collector. Typical design: emitter heavily doped; base lightly doped and very thin. Active region biasing: emitter–base junction forward biased; collector–base junction reverse biased. Concept / Approach: In a p–n–p transistor, the main injection from the emitter into the base is of holes (majority carriers of the p-emitter). Some counter-injection of electrons from the n-base back into the emitter exists, but good design ensures it is much smaller. Hence IEp (holes) is much larger than IEn (electrons), making emitter efficiency close to unity. Step-by-Step Solution: Recognize emitter efficiency γ ≈ IEp / IE and is designed to be high.Heavily dope the emitter (p-type) to maximize hole injection; lightly dope the base to minimize electron back-injection.Thus IEp >> IEn under normal active operation. Verification / Alternative check: Mirror the logic for n–p–n devices: there electron injection dominates (from n-emitter), reinforcing the general rule that the emitter injects its majority carriers predominantly. Why Other Options Are Wrong: (a) “Almost equal” contradicts the purpose of heavy emitter doping and thin, lightly doped base.(c) Reverses the actual dominance.(d) Ambiguous; correct device physics picks a clear dominance.(e) IEp cannot be ≈ 0 because holes from the p-emitter are the very cause of conduction. Common Pitfalls: Confusing minority vs majority injection and ignoring doping profiles. Final Answer: IEp >> IEn

Question 4

In the photoelectric effect, on which parameter does the amount of photoelectric emission current primarily depend, assuming the frequency is above threshold?

Accepted Answer

Introduction / Context: The photoelectric effect underpins modern optoelectronics and quantum theory. Quantities like photocurrent, stopping potential, and kinetic energy of emitted electrons each depend on different aspects of the incoming light. This question distinguishes the role of intensity versus frequency in setting the magnitude of the emitted current. Given Data / Assumptions: Photon energy E = h f; emission occurs only if f ≥ f0 (threshold frequency). Once f > f0, electrons are emitted. Photocurrent is proportional to the number of emitted electrons per unit time reaching the anode in a circuit. Concept / Approach: For f ≥ f0, the number of photons arriving per second determines how many electrons are emitted (assuming each photon can liberate at most one photoelectron). Intensity sets photon flux: higher intensity → more photons per second → larger photoelectric current. Frequency determines the maximum kinetic energy of photoelectrons (via stopping potential), not the magnitude of current, provided a sufficient collection field is applied and f remains above threshold. Step-by-Step Solution: Ensure f ≥ f0 so emission can occur (else current is zero regardless of intensity).Relate intensity I to photon flux Nγ: Nγ ∝ I / (h f).Photocurrent Iph ∝ electron emission rate ∝ Nγ, hence Iph ∝ intensity. Verification / Alternative check: Stopping potential Vstop is set by electron kinetic energy KEmax = h f − φ (φ is work function). Varying frequency shifts KE but not directly the current magnitude if the collection field is adequate. Why Other Options Are Wrong: Frequency alone: governs KE, not emission rate (above threshold).Both frequency and intensity: frequency must exceed threshold, but the amount of current primarily scales with intensity under typical measurement conditions.Work function only: sets threshold, not current magnitude at a fixed intensity. Common Pitfalls: Confusing stopping potential measurements (frequency-dependent) with photocurrent magnitude (intensity-dependent). Final Answer: intensity of incident radiation

Question 5

At room temperature, in an intrinsic (pure) semiconductor, which charge carriers are responsible for electrical conduction?

Accepted Answer

Introduction / Context: Intrinsic semiconductors (e.g., undoped silicon) exhibit thermally generated carriers even without intentional doping. Recognizing that both electron and hole populations participate in conduction is foundational for understanding conductivity, intrinsic carrier concentration, and the effect of temperature on devices. Given Data / Assumptions: Intrinsic semiconductor with bandgap Eg. At room temperature, thermal energy produces electron–hole pairs (EHPs). Carrier concentrations are equal: n = p = ni. Concept / Approach: Every thermally generated electron in the conduction band leaves behind a hole in the valence band. Thus, intrinsic conduction is a combination of electron drift/diffusion and hole drift/diffusion. Conductivity is σ = q(n μn + p μp) = q ni (μn + μp). Both carrier types contribute concurrently to current flow under an applied electric field. Step-by-Step Solution: Recognize that generation creates paired carriers: electrons and holes.Use intrinsic condition: n = p = ni; both terms appear in σ.Therefore, total current has components from both electrons and holes. Verification / Alternative check: If the material were doped (n-type or p-type), majority carriers dominate, but in the intrinsic case the symmetry between n and p confirms the answer. Why Other Options Are Wrong: Holes only or electrons only ignores the paired nature of intrinsic generation.Ions/excitons: not the principal current carriers in standard intrinsic conduction under bias. Common Pitfalls: Assuming one carrier dominates because μn > μp; although mobilities differ, both carrier densities are equal and both contribute. Final Answer: holes and electrons

Question 6

In the LED protection circuit (series resistor R with the LED, and a diode D connected to protect the LED), what are the functions of R and D respectively?

Accepted Answer

Introduction / Context: LEDs are current-driven devices with a relatively fixed forward drop over a small current range. Practical LED circuits therefore include a series resistor to set current and a protective diode across the LED (usually antiparallel) to guard against reverse voltage during polarity reversals or inductive transients. Given Data / Assumptions: Series resistor R is present with the LED for current control. A diode D is wired in antiparallel (or an equivalent protective orientation) to the LED. Reverse breakdown of typical indicator LEDs is low (often 5 V or less), so reverse stress must be limited. Concept / Approach: The LED must be protected from excessive forward current and from reverse breakdown. The resistor limits forward current: I ≈ (Vs − Vf)/R. The protection diode D provides a safe reverse path with low drop, clamping the reverse voltage across the LED to approximately a diode drop, thereby preventing reverse breakdown of the LED junction. Step-by-Step Solution: Forward conduction: R sets I_LED to a safe value by Ohm’s law.Reverse polarity or transient: D conducts first, keeping Vreverse across LED ≈ 0.7–1 V rather than several volts.Therefore, R limits current; D protects against reverse breakdown. Verification / Alternative check: Check LED datasheet: maximum reverse voltage ratings are low; external diode is recommended in AC drive or inductive environments. Why Other Options Are Wrong: R does not “limit voltage” directly; it is a current-limiting element.D does not protect primarily against forward “over-current”; that is handled by R.“None” and “boost current” options contradict safe design practice. Common Pitfalls: Omitting the antiparallel diode when driving LEDs from AC or when inductive kick can occur. Final Answer: to limit the current and protect LED against reverse breakdown voltage.

Question 7

In a bipolar junction transistor (BJT), which terminal current is the largest under normal operation, and how are the three currents related?

Accepted Answer

Introduction / Context: Kirchhoff’s current law (KCL) applied to the transistor junctions provides a direct relation among emitter (IE), base (IB), and collector (IC) currents. Recognizing that IE is the sum of the other two is vital for quick bias calculations and understanding of current gain definitions α and β. Given Data / Assumptions: Steady-state operating point in any BJT configuration. KCL at the transistor: IE = IB + IC. Typically, IB ≪ IC (for a transistor with reasonable β), hence IE is marginally larger than IC. Concept / Approach: Since the emitter injects carriers that split into the collector (mostly) and the base (small recombination), IE is necessarily the largest terminal current. With β = IC/IB high (e.g., 100), IB is small; IC is slightly less than IE by the amount of IB. Therefore ordering is IE > IC ≫ IB. Step-by-Step Solution: Write KCL: IE = IC + IB.Given IB is small, IE ≈ IC + a small increment, making IE the largest.Hence, the correct selection is “emitter current”. Verification / Alternative check: For β = 100 and IC = 10 mA, IB = 0.1 mA, IE = 10.1 mA → IE largest, confirming the relation. Why Other Options Are Wrong: Collector current is slightly less than IE by IB.Base current is the smallest in a well-designed BJT bias.“Either … or …” contradicts the fixed KCL relation. Common Pitfalls: Forgetting KCL or misinterpreting β and α, leading to incorrect ordering of currents. Final Answer: emitter current

Question 8

Assertion–Reason: (A) The conductivity of a p-type semiconductor is higher than that of an intrinsic semiconductor. (R) Adding donor impurities creates extra energy levels just below the conduction band.

Accepted Answer

Introduction / Context: Assertion–Reason questions evaluate not only factual correctness but also causal linkage. Here we compare conductivity of p-type versus intrinsic material and examine whether the stated reason logically supports the assertion. Given Data / Assumptions: Intrinsic semiconductor: n = p = ni; conductivity σi = q ni (μn + μp). p-type semiconductor: majority holes with p ≫ n; conductivity σp ≈ q p μp with p set by acceptor doping. Donor impurities (n-type) introduce energy levels near the conduction band; acceptor impurities (p-type) introduce levels near the valence band. Concept / Approach: The assertion states p-type conductivity exceeds intrinsic conductivity—usually true, because doping greatly increases majority carrier concentration compared to ni at room temperature. The reason, however, mentions donor impurities (n-type mechanism) rather than acceptors, so it does not explain the improved conductivity of a p-type semiconductor. Thus A is true, R is false in this context. Step-by-Step Solution: Assess A: Doping (acceptors) raises hole concentration well above ni → σ increases → A is true.Assess R: Donor levels below conduction band correspond to n-type doping, irrelevant to p-type improvement → R is false as an explanation for A.Therefore, correct choice: A true, R false. Verification / Alternative check: If R had referred to acceptor levels near the valence band (promoting holes), it could have explained A. Since it does not, the pairing fails. Why Other Options Are Wrong: (a) Incorrect because R does not support A.(b) R is not even true in the p-type context used to explain A.(d) A is not false; p-type typically has higher σ than intrinsic. Common Pitfalls: Overlooking the specific impurity type (donor vs acceptor) in reason statements. Final Answer: A is true but R is false

Question 9

For normal transistor operation, how should the two p–n junctions of a BJT be biased to achieve the active region?

Accepted Answer

Introduction / Context: The BJT has two junctions: emitter–base (E–B) and collector–base (C–B). The mode of operation (cutoff, active, saturation, reverse-active) is determined by the biasing of these junctions. Correct biasing is essential for amplification in the active region. Given Data / Assumptions: Active region: transistor used as a linear amplifier. E–B junction: forward biased to inject carriers from emitter into base. C–B junction: reverse biased to collect injected carriers into the collector. Concept / Approach: In active operation, the emitter injects majority carriers into the base (forward bias). The collector–base junction reverse bias creates a strong field that sweeps the minority carriers across the depletion region into the collector, enabling controlled, nearly linear amplification governed by the base current (or base–emitter voltage in transconductance terms). Step-by-Step Solution: Set VEB forward (≈ 0.7 V for Si, ≈ 0.3 V for Ge).Set VCB reverse (positive for n–p–n, negative for p–n–p collector with respect to base in conventional bias).This biasing yields the active region necessary for amplification. Verification / Alternative check: Compare other modes: both forward → saturation; both reverse → cutoff; swapped (reverse-active) rarely used for amplification. Why Other Options Are Wrong: Both forward/reverse bias correspond to saturation/cutoff respectively, not linear amplification.Specified “forward at C–B and reverse at E–B” describes reverse-active mode. Common Pitfalls: Confusing collector polarity with bias type; always think in terms of E–B and C–B junctions. Final Answer: one junction is forward biased and the other is reverse biased

Question 10

A BJT shows a current gain α = 0.99 in common-base (CB) configuration. What is the corresponding current gain in common-collector (CC, emitter follower) configuration?

Accepted Answer

Introduction / Context: Current gains in different BJT configurations are related. Converting between α (CB gain) and β (CE gain) enables quick estimates of follower (CC) current gain, which is approximately (β + 1). Understanding these relations helps in rapid hand analysis and design sanity checks. Given Data / Assumptions: Common-base current gain α = IC/IE = 0.99. Relation between α and β: β = α / (1 − α). Common-collector current gain (approx.) = β + 1 (ratio IE/IB). Concept / Approach: First compute β from α, then obtain the CC gain. A high α implies a high β, and consequently a CC current gain near β + 1, often close to 1 for voltage gain but high for current gain, which is the essence of the emitter follower behavior. Step-by-Step Solution: Compute β: β = α/(1 − α) = 0.99 / 0.01 = 99.Compute CC current gain: ≈ β + 1 = 100.Therefore the current gain in CC configuration is 100. Verification / Alternative check: Check that IE = IC + IB and IC ≈ α IE; with α = 0.99, IB is small, giving a large IE/IB ratio ~ 100. Why Other Options Are Wrong: 99 overlooks the +1 term in CC gain.1.01 or 0.99 confuse current gain with voltage gain (which is near unity in CC).“~50” is arbitrary and inconsistent with α = 0.99. Common Pitfalls: Mixing up current gain (high for CC) with voltage gain (≈ 1) of an emitter follower. Final Answer: 100

Question 11

A p-n junction diode has

Accepted Answer

A p-n Junction has all these features.

Question 12

Assertion (A): The reverse saturation current in a semiconductor diode is 4nA at 20°C and 32 nA at 50°C. Reason (R): The reverse saturation current in a semiconductor diode doubles for every 10°C rise in temperature.

Accepted Answer

At 20°C, 4 nA, at 30°C, 8 nA, at 40°C, 16 nA, at 50°C, 32 nA.

Question 13

Calculate the stability factor and change in IC from 25°C to 100°C for, β = 50, RB/ RE = 250, ΔIC0 = 19.9 nA for emitter bias configuration.

Accepted Answer

SICO = (1 + β). ⟹ 51. = 42.53 ΔIC = (SICO).ΔICO = 42.53 x 19.9 nA = 0.85 μA.

Question 14

An increase in temperature increases the width of depletion layer.

Accepted Answer

With increase in temperature width of depletion layer decreases.

Question 15

An amplifier without feedback has a voltage gain of 50, input resistance of 1 kΩ and output resistance of 2.5 kΩ. The input resistance of the current shunt -ve feedback amplifier using the above amplifier with a feedback factor of 0.2 is

Accepted Answer

Input Resistance with feedback for current shunt, .

Question 16

Work Function in Metals: Determine whether the statement "Work function is the maximum energy required by the fastest electron at 0 K to escape from the metal surface" is true or false.

Accepted Answer

Introduction / Context: The concept of work function is fundamental in solid-state physics and electronics. It governs electron emission phenomena such as thermionic emission, photoelectric effect, and field emission. Understanding its correct definition ensures clarity when analyzing device physics like cathodes, photocells, and semiconductor-metal contacts. Given Data / Assumptions: Material is a pure metal at absolute zero (0 K). Statement: "Work function is the maximum energy required by the fastest electron at 0 K to escape from the metal surface." Concept / Approach: The work function (Φ) is defined as the minimum energy required to remove an electron from the Fermi level of a metal to a point just outside the surface (vacuum level). At 0 K, the highest-energy electrons occupy the Fermi level. Thus, the energy difference between the vacuum level and the Fermi level represents the work function. Step-by-Step Solution: At 0 K, electrons fill energy states up to the Fermi energy (E_F).The vacuum energy level (E_vac) is above E_F.Work function Φ = E_vac − E_F.Therefore, it is indeed the maximum energy required by the fastest electron (at the Fermi level) at 0 K to escape. Verification / Alternative check: Photoelectric emission experiments confirm that the threshold photon energy equals the work function. At 0 K, the fastest electrons are at E_F, so the definition matches practical measurements. Why Other Options Are Wrong: 'False': Contradicts the well-established definition. 'True only for semiconductors': Misleading; definition applies to metals and semiconductors alike. 'False, depends on temperature': Temperature influences electron population above E_F but does not redefine Φ. 'Cannot be determined': Incorrect because the definition is standard. Common Pitfalls: Confusing 'minimum energy required' with 'maximum electron energy available' — at 0 K, these coincide because only electrons at the Fermi level can escape. Believing that work function changes at 0 K — while thermal effects exist, the base definition is independent of operating temperature. Final Answer: True

Question 17

Crossover Distortion in Amplifiers: Identify which amplifier class typically exhibits crossover distortion behavior.

Accepted Answer

Introduction / Context: Crossover distortion is a common non-linearity in amplifier circuits. It appears near the zero-crossing of the input signal, particularly in push–pull output stages. Understanding which class of amplifier exhibits this distortion helps in selecting proper biasing techniques. Given Data / Assumptions: Four amplifier classes given: A, B, AB, and an unrelated term 'common pulse.' We want to identify which one characteristically shows crossover distortion. Concept / Approach: Class B amplifiers conduct for exactly 180° of the input cycle per transistor. Near the zero crossing, both transistors are off briefly due to base–emitter voltage requirements, leading to distortion called 'crossover distortion.' Step-by-Step Solution: Class A: Device conducts for full 360°, highly linear, no crossover distortion.Class B: Each device conducts for 180°, gap around zero crossing causes crossover distortion.Class AB: Devices conduct for slightly more than 180°, reducing but not eliminating crossover distortion.Class C: Not listed in the options but typically conducts less than 180°, unsuitable for linear amplification. Verification / Alternative check: Practical push–pull Class B designs exhibit crossover distortion unless biasing (Class AB) is applied to minimize it. Audio engineers frequently discuss this issue in amplifier design. Why Other Options Are Wrong: Class A: No crossover distortion. Class AB: Reduces distortion significantly but not characteristic. 'Common pulse' is irrelevant and incorrect. Common Pitfalls: Confusing harmonic distortion (common in Class C) with crossover distortion (specific to Class B). Assuming all amplifier classes exhibit it equally. Final Answer: class B O/P stage

Question 18

Reverse Recovery Time in Diodes: Identify which type of diode exhibits nearly zero reverse recovery time.

Accepted Answer

Introduction / Context: Reverse recovery time is a key performance parameter in diodes, particularly in high-frequency or fast-switching applications. It refers to the time taken for the diode to stop conducting after switching from forward bias to reverse bias. Some diodes are designed to minimize or nearly eliminate this delay. Given Data / Assumptions: We are comparing Zener, Tunnel, Schottky, and PIN diodes. Focus is on reverse recovery time, not other characteristics like breakdown voltage or capacitance. Concept / Approach: Reverse recovery occurs due to stored charge carriers in the depletion or neutral regions. If a diode conducts by majority carriers only (no storage), reverse recovery is nearly zero. Schottky diodes are majority-carrier devices (metal–semiconductor junction), so they exhibit negligible reverse recovery time. Step-by-Step Solution: Zener diode: Reverse-biased breakdown operation, not optimized for fast switching.Tunnel diode: Quantum tunneling conduction, not designed for switching with low recovery.PIN diode: Large intrinsic region, significant stored charge, hence longer recovery.Schottky diode: No minority carrier storage → negligible recovery time. Verification / Alternative check: Switching tests confirm Schottky diodes outperform all listed options in terms of recovery speed. They are widely used in RF, rectification, and power electronics where fast switching is critical. Why Other Options Are Wrong: Zener: Recovery time dominated by avalanche processes. Tunnel: Operates differently but still not characterized by near-zero recovery. PIN: Designed for RF attenuation, not ultrafast recovery. Common Pitfalls: Confusing breakdown diodes with fast-switching diodes. Assuming all 'special' diodes are fast; only Schottky uniquely offers near-zero recovery. Final Answer: Schottky diode

Question 19

FET Small-Signal Amplification: Based on the V–I characteristics of a Field Effect Transistor (FET), identify the region in which the device should be biased for small-signal amplification.

Accepted Answer

Introduction / Context: For FET amplifiers, proper biasing ensures linear operation with good gain. The device characteristics (drain current vs. drain–source voltage) show several distinct regions: cutoff, ohmic (linear), saturation (active), and breakdown. Amplification is only feasible in one of them. Given Data / Assumptions: Typical n-channel JFET or MOSFET characteristic curve is implied. Regions: AB (cutoff), BC (ohmic), CD (saturation/active), BD (includes breakdown). Concept / Approach: The small-signal amplification region corresponds to the saturation (also called active) region, where drain current is controlled primarily by gate–source voltage and remains nearly constant over a range of drain–source voltages. Step-by-Step Solution: Cutoff region (AB): No conduction; cannot amplify.Ohmic region (BC): Acts as a variable resistor, not suitable for linear voltage amplification.Saturation region (CD): ID depends on VGS; current is nearly constant w.r.t. VDS, allowing linear amplification.Breakdown (BD): Device fails; not a usable region. Verification / Alternative check: Practical FET amplifier bias points are always chosen in the saturation region, validating CD as the correct choice. Why Other Options Are Wrong: AB: Cutoff, no current. BC: Resistive behavior, not linear gain. BD: Beyond safe operation. Common Pitfalls: Confusing MOSFET digital switching (cutoff & saturation in digital sense) with analog small-signal operation (saturation = active region). Final Answer: CD

Question 20

CE Transistor Regions of Operation: In which region of a common-emitter (CE) bipolar junction transistor is the collector current almost constant?

Accepted Answer

Introduction / Context: Bipolar junction transistors (BJTs) operate in distinct regions: cutoff, active, saturation, and breakdown. Recognizing which region provides linear amplification with nearly constant collector current is essential in amplifier circuit design. Given Data / Assumptions: Configuration: Common-emitter (CE). Focus is on collector current (IC) behavior with varying collector–emitter voltage (VCE). Concept / Approach: In the active region, IC is controlled primarily by the base current (IB) or base–emitter voltage (VBE), and remains nearly constant over a wide range of VCE. This property makes the active region suitable for amplification. In contrast, in saturation IC depends strongly on VCE, and in breakdown IC rises uncontrollably. Step-by-Step Solution: Cutoff: IB = 0, IC ≈ 0.Active: IC ≈ β * IB, nearly constant over VCE changes (until saturation or breakdown limits).Saturation: IC increases strongly with VCE, not constant.Breakdown: IC skyrockets, not constant. Verification / Alternative check: Characteristic IC–VCE curves for BJTs show flat regions (active) where IC is almost constant, validating this answer. Why Other Options Are Wrong: Saturation: IC depends on VCE strongly. Breakdown: Destructive runaway. Both saturation and active: Misleading, only active region shows flat IC vs VCE behavior. Cutoff: No conduction at all. Common Pitfalls: Confusing 'digital saturation' (switch closed) with 'linear constant current' behavior. Assuming collector current is constant in both saturation and active regions; in reality only active qualifies. Final Answer: Active region

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